Question: Simplify and expand the following expression: $ \dfrac{3}{3z + 6}- \dfrac{2}{3z - 27}+ \dfrac{4}{z^2 - 7z - 18} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the first term: $ \dfrac{3}{3z + 6} = \dfrac{3}{3(z + 2)}$ We can factor a $3$ out of denominator in the second term: $ \dfrac{2}{3z - 27} = \dfrac{2}{3(z - 9)}$ We can factor the quadratic in the third term: $ \dfrac{4}{z^2 - 7z - 18} = \dfrac{4}{(z + 2)(z - 9)}$ Now we have: $ \dfrac{3}{3(z + 2)}- \dfrac{2}{3(z - 9)}+ \dfrac{4}{(z + 2)(z - 9)} $ The least common multiple of the denominators is: $ 9(z + 2)(z - 9)$ In order to get the first term over $9(z + 2)(z - 9)$ , multiply by $\dfrac{3(z - 9)}{3(z - 9)}$ $ \dfrac{3}{3(z + 2)} \times \dfrac{3(z - 9)}{3(z - 9)} = \dfrac{9(z - 9)}{9(z + 2)(z - 9)} $ In order to get the second term over $9(z + 2)(z - 9)$ , multiply by $\dfrac{3(z + 2)}{3(z + 2)}$ $ \dfrac{2}{3(z - 9)} \times \dfrac{3(z + 2)}{3(z + 2)} = \dfrac{6(z + 2)}{9(z + 2)(z - 9)} $ In order to get the third term over $9(z + 2)(z - 9)$ , multiply by $\dfrac{9}{9}$ $ \dfrac{4}{(z + 2)(z - 9)} \times \dfrac{9}{9} = \dfrac{36}{9(z + 2)(z - 9)} $ Now we have: $ \dfrac{9(z - 9)}{9(z + 2)(z - 9)} - \dfrac{6(z + 2)}{9(z + 2)(z - 9)} + \dfrac{36}{9(z + 2)(z - 9)} $ $ = \dfrac{ 9(z - 9) - 6(z + 2) + 36} {9(z + 2)(z - 9)} $ Expand: $ = \dfrac{9z - 81 - 6z - 12 + 36}{9z^2 - 63z - 162} $ $ = \dfrac{3z - 57}{9z^2 - 63z - 162}$ Simplify: $ = \dfrac{z - 19}{3z^2 - 21z - 54}$